Simplify and expand the following expression: $ \dfrac{4}{3x - 30}+ \dfrac{1}{x + 3}- \dfrac{4}{x^2 - 7x - 30} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{4}{3x - 30} = \dfrac{4}{3(x - 10)}$ We can factor the quadratic in the third term: $ \dfrac{4}{x^2 - 7x - 30} = \dfrac{4}{(x - 10)(x + 3)}$ Now we have: $ \dfrac{4}{3(x - 10)}+ \dfrac{1}{x + 3}- \dfrac{4}{(x - 10)(x + 3)} $ The least common multiple of the denominators is: $ 3(x - 10)(x + 3)$ In order to get the first term over $3(x - 10)(x + 3)$ , multiply by $\dfrac{x + 3}{x + 3}$ $ \dfrac{4}{3(x - 10)} \times \dfrac{x + 3}{x + 3} = \dfrac{4(x + 3)}{3(x - 10)(x + 3)} $ In order to get the second term over $3(x - 10)(x + 3)$ , multiply by $\dfrac{3(x - 10)}{3(x - 10)}$ $ \dfrac{1}{x + 3} \times \dfrac{3(x - 10)}{3(x - 10)} = \dfrac{3(x - 10)}{3(x - 10)(x + 3)} $ In order to get the third term over $3(x - 10)(x + 3)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{4}{(x - 10)(x + 3)} \times \dfrac{3}{3} = \dfrac{12}{3(x - 10)(x + 3)} $ Now we have: $ \dfrac{4(x + 3)}{3(x - 10)(x + 3)} + \dfrac{3(x - 10)}{3(x - 10)(x + 3)} - \dfrac{12}{3(x - 10)(x + 3)} $ $ = \dfrac{ 4(x + 3) + 3(x - 10) - 12} {3(x - 10)(x + 3)} $ Expand: $ = \dfrac{4x + 12 + 3x - 30 - 12}{3x^2 - 21x - 90} $ $ = \dfrac{7x - 30}{3x^2 - 21x - 90}$